Going from u to v or from v to u? Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19234   Accepted: 5182 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors…

Going from u to v or from v to u? Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15494   Accepted: 4100 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors…

2013-09-08 10:00 var m, n :longint; t :longint; f, last :..] of longint; pre, other :..] of longint; l, time :longint; dfn, low :..] of longint; tot :longint; stack :..] of longint; flag, fs :..] of boolean; i :longint; key :..] of longint; kk :longi…

这题搞了好久,先是拓扑排序这里没想到,一开始自己傻乎乎的跑去找每层出度为1的点,然后才想到能用拓扑排序来弄. 拓扑排序的时候也弄了挺久的,拓扑排序用的也不多. 题意:给一个图求是否从对于任意两个点能从v 到w 或者从w到v连通. 思路:单连通,先强连通缩点,若scnt为1,或者出度为零的点为0,直接输出YES,若出度为零的点大于1,则代表有分支输出NO.若出度为零的点为1,判断组成的树是否为单链,即没有分支,用拓扑排序即可. 代码: #include<iostream> #include<…

Going from u to v or from v to u? Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17383   Accepted: 4660 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors…

好久没写过这么长的代码了,题解东哥讲了那么多,并查集优化还是很厉害的,赶快做做前几天碰到的相似的题. #include <iostream> #include <algorithm> #include <cstdio> using namespace std; , M = 2e5 + ; ], Next[M * ]; int dfn[N], low[N], n, m, tot, num; ]; void add(int x, int y) { ver[++tot] =…

Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 37111   Accepted: 15124 Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M &…

Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 37083   Accepted: 15104 Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M &…

题目大意: 在一个有向图中,每两点间通信需要一定的时间,但同一个强连通分量里传递信息不用时间,给两点u,v求他们最小的通信时间.   解题过程: 1.首先把强连通分量缩点,然后遍历每一条边来更新两个强联通分量之间的距离.. 2.直接Floyd会超时,应该用dijstra或者spfa做k次最短路.   犯的错误:前向星数组开的太小,一直超时.  …

Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16806   Accepted: 6643 Description A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a li…